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I'm looking for the Fourier transform of $f(t)=\frac{1}{t}\theta(t)$ ($\theta$ is the step function), I know how to do both factors separately but not if they are multiplying. Can someone help me ?

I try to do that in the wolframalpha page but it couldn't (at least by free).

And I'm keen on the Fourier transform of $e^{-\frac{1}{t}}\theta(t)$ or $e^{-\frac{1}{t^2}}\theta(t)$ too, actually I'm more interested in this two last ones than in $\hat{f}$.

Many thanks in advance.

Gabriel Palau
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    These computations and more are in vol 1 of Gelfand-shilov on generalized functions. You might also want to clarify what $f(t)$ is. It should be a temperate distribution but it is not locally integrable, so you probably meant some principal value. – Abdelmalek Abdesselam Jan 30 '19 at 17:23

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You cannot multiply distributions, you have to define $\theta(t)/t$ separately. If it's defined as $$\left( \frac {\theta(t)} t, \phi \right) = \int_{t > 0} \frac {\phi(t) - \phi(0) [t < 1]} t dt,$$ then you can combine known results for the transforms of $1/t$ and $1/|t|$ to obtain $$\left( \frac {\theta(t)} t, e^{i p t} \right) = -\ln(-i p ) - \gamma.$$

For the other two functions, these integrals exist in the ordinary sense: $$\int_0^\infty (e^{-1/t} - 1) \,e^{i p t} dt = \frac {2 K_1(2 \sqrt{-i p \,})} {\sqrt{-i p \,}} -\frac i p, \\ \int_0^\infty (e^{-1/t^2} - 1) \,e^{i p t} dt = \cases{ \frac i {p \sqrt \pi} G_{0, 3}^{3, 0} \left( -\frac {p^2} 4 \middle| {- \atop 0, \frac 1 2, 1} \right) -\frac i p, & $p < 0$ \\ \\ \frac i {p \sqrt \pi} G_{3, 0}^{0, 3} \left( -\frac 4 {p^2} \middle| {0, \frac 1 2, 1 \atop -} \right) -\frac i p, & $p > 0$},$$ where $K$ is the modified Bessel function and $G$ is the Meijer G-function. Then adding the transform of $\theta(t)$ gives the answer.

Maxim
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  • Did you miss a $\delta(p)/2$ ? Pick any function with $\psi(0) = 1$ and let $(f_\psi,\phi) = (sign(t)/t, \phi-\phi(0) \psi)$ then $( f_\psi,t,\phi) = (sign(t), \phi)$ so $\hat{f_\psi}' = i \widehat{sign}(p)= pv(\frac{-1}{p}), \hat{f_\psi} =- \ln |p| + C_\psi$ and with $g_\psi = f_\psi + 1/2$ then $\hat{g_\psi} =- \ln |p| + C_\psi+\delta(p)/2$ – reuns Jan 31 '19 at 17:42
  • The last step seems to be the issue: $f_\psi$ is a regularization of the ordinary function $1/|t|$, but $f_\psi + 1/2$ is not a regularization of $\theta(t)/t$. – Maxim Jan 31 '19 at 19:45
  • Right I meant $g_\psi = f_\psi+pv(1/2t)$, $\hat{g_\psi} = -\ln |p| +C_\psi-i\pi sign(p)$ which is compatible with your result – reuns Jan 31 '19 at 20:10